由f1(x)=0∉(0,1)得f1(x)在D上不封闭.f2(x)=-x2-x+1在(0,1)上是减函数,0=f2(1)<f2(x)<f2(0)=1,得f2(x)适合.而f3(x)=1-x在(0,1)上是减函,0=f3(1)<f3(x)<f3(0)=1,f3(x)适合.而f4(x)=x在(0,1)上是增函数,且0=f4(0)<f4(x)<f4(1)=1,f4(x)适合.从而得到结果.
【解析】
∵f1=0∉(0,1),
∴f1(x)在D上不封闭.
∵f2(x)=-x2-x+1在(0,1)上是减函数,
∴0=f2(1)<f2(x)<f2(0)=1,
∴f2(x)适合.
∵f3(x)=1-x在(0,1)上是减函数,
∴0=f3(1)<f3(x)<f3(0)=1,
∴f3(x)适合.
又∵f4(x)=x在(0,1)上是增函数,
且0=f4(0)<f4(x)<f4(1)=1,
∴f4(x)适合.
故答案为:②③④